imc

Let $l_1$ be the line containing $A$ and $B$ and let $l_2$ be the line containing $C$ and $D$. If we set the bottom left point at $(0,0)$, then $A=(0,3)$, $B=(6,0)$, $C=(4,2)$, and $D=(2,0)$. The line $l_1$ is given by the equation $y=m_{1}x+b_1$. The $y$-intercept is $A=(0,3)$, so $b_1=3$. We are given two points on $l_1$, hence we can compute the slope, $m_1$ to be $\frac{0-3}{6-0}=-\frac{1}{2}$, so $l_1$ is the line $y=\frac{-1}{2}x+3$ Similarly, $l_2$ is given by $y=m_{2}x+b_2$. The slope in this case is $\frac{2-0}{4-2}=1$, so $y=x+b_2$. Plugging in the point $(2,0)$ gives us $b_2=-2$, so $l_2$ is the line $y=x-2$. At $E$, the intersection point, both of the equations must be true, so $$\begin{array}{rl}y=x-2,y=\frac{-1}{2}x+3& \Rightarrow x-2=\frac{-1}{2}x+3\\ & \Rightarrow x=\frac{10}{3}\\ & \Rightarrow y=\frac{4}{3}\end{array}$$ We have the coordinates of $A$ and $E$, so we can use the distance formula here: $$\sqrt{{\left(\frac{10}{3}-0\right)}^2+{\left(\frac{4}{3}-3\right)}^2}=\frac{5\sqrt{5}}{3}$$ which is answer choice $\boxed{\frac{5\sqrt{5}}{3}}$